Difference of two squares \( (a+b)(a-b)=a^{2}-b^{2} \) \( (c-d)(c+d)=c^{2}-d^{2} \) 1. \( (x-2)(x-2) \) 2. \( (x-a)(x+a) \) 3. \( (3 x+4)(3 x-4) \) 4 \( [x-(a-4)][x+(a-4)] \)

Certainly! Let's apply the **difference of two squares** formula to each of the given expressions. The difference of two squares is given by: \[ (a + b)(a - b) = a^2 - b^2 \] Now, let's examine each problem: --- ### **1. \( (x - 2)(x - 2) \)** **Interpretation:** This is actually a **perfect square** rather than a difference of squares because both binomials are the same. **Expansion:** \[ (x - 2)^2 = x^2 - 4x + 4 \] --- ### **2. \( (x - a)(x + a) \)** **Application of Difference of Squares:** Here, \( a = x \) and \( b = a \). \[ (x - a)(x + a) = x^2 - a^2 \] **Result:** \[ x^2 - a^2 \] --- ### **3. \( (3x + 4)(3x - 4) \)** **Application of Difference of Squares:** Here, \( a = 3x \) and \( b = 4 \). \[ (3x + 4)(3x - 4) = (3x)^2 - 4^2 = 9x^2 - 16 \] **Result:** \[ 9x^2 - 16 \] --- ### **4. \( [x - (a - 4)][x + (a - 4)] \)** **Simplify the Expression:** First, simplify the binomials inside the brackets. \[ [x - (a - 4)] = (x - a + 4) \\ [x + (a - 4)] = (x + a - 4) \] **Application of Difference of Squares:** Let \( c = x \) and \( d = a - 4 \). \[ (x - (a - 4))(x + (a - 4)) = x^2 - (a - 4)^2 \] **Expand the Square:** \[ x^2 - (a^2 - 8a + 16) = x^2 - a^2 + 8a - 16 \] **Result:** \[ x^2 - a^2 + 8a - 16 \] --- **Summary of Results:** 1. \( (x - 2)^2 = x^2 - 4x + 4 \) 2. \( (x - a)(x + a) = x^2 - a^2 \) 3. \( (3x + 4)(3x - 4) = 9x^2 - 16 \) 4. \( [x - (a - 4)][x + (a - 4)] = x^2 - a^2 + 8a - 16 \) Feel free to ask if you need further clarification on any of these steps!